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The pH of 0.02 M NH(4)Cl (aq) (pK(b)=4.7...

The `pH` of `0.02 M NH_(4)Cl (aq) (pK_(b)=4.73)` is equal to

A

`3.78`

B

`4.73`

C

`5.48`

D

`7.00`

Text Solution

Verified by Experts

The correct Answer is:
C
`pH = 1/2 [pK_(w) -pK_(b) - logc] `
` = 1/2 [ 14 - 4.73 - log 0.02 ] `
`1/2 [ 14 - 4.73 + 1.698 ] = 5.48`
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